Connected metric spaces with at least 2 points are uncountable. That's a problem I proved (quite a while back) in tiny Rudin. However, I don't really get it. The other questions were actually useful results - I don't think I've ever come near using this result. Surely it's going to be close to apparent that you're working in an uncountable set?
For instance, examples where this result could be applied but it is hard otherwise to tell that the space is uncountable?
 A: A mildly interesting remark:
A comment suggests this result can be used to prove the uncountability of $R$.  But the proofs here and elsewhere of this result typically use the existence of a surjective function to an interval $[a,b]$ and the uncountability of that interval. So the argument suggested in the comment would be circular.  Here, however, is a nice proof of this result that does not use uncountability of $R$.
Let $X = \{x_n\}_{n=1}^\infty$ be a connected metric space and suppose
$\epsilon >0$. Let $U_n$ be the open ball of radius $\epsilon/2^n$ centered at $x_n$.  Define a relation ~ on $X$ by $x$ ~ $y$ if there is a sequence ${n_i}$, $1 \le i \le k$ such that $x \in U_{n_1}$, $y \in U_{n_k}$ and $U_{n_i} \cap U_{n_{i+1}} \ne \emptyset.$  Let $z_{n_i}$
be a point in $U_{n_i} \cap U_{n_{i+1}}$
It is immediate that ~ is an equivalence relation and that equivalence classes are open sets.  Since $X$ is connected there can be only one equivalence class. It is also easy to see that $x$ ~ $y$ implies
$$d(x,y) < 2\sum_{i=1}^{k-1}d(x_{n_i}, x_{n_{i+1}}) < 
2\sum_{i=1}^{k-1}(d(x_{n_i}, z_{n_{i}}) + d(z_{n_i}, x_{n_{i+1}}))<4\sum_{i=1}^k
\frac{\epsilon}{2^{n_i}} < 4\epsilon. $$
Since $x,y$ and $\epsilon >0$ are arbitrary, it follows that the diameter of $X$ is $0$ and $X$ is a singleton.
Another "easy" proof of uncountability of $R$ uses the fact that the Lebesgue measure of an interval is positive and countable sets have measure zero.  But the common proof of fact that the measure of a non-trivial interval is positive uses connectivity and is similar to the argument above.  I think if one tries to dig into the topological roots of the measure theory proof you end up with something like the argument above.
A: Let your points be $a$ and $b$. 
Let $\lambda\in(0,1)$ Suppose there is no point $x$ in your space such that $d(a,x)=\lambda d(a,b)$. Then the sets $\{z:d(a,z)<\lambda d(a,b)\}$ and $\{z:d(a,z)>\lambda d(a,b)\}$ are two non-empty open sets which partition the space.
Since we are assuming connectedness, this is impossible.
Therefore, the image of the function $d(a,\mathord\cdot):X\to\mathbb R$ contains the interval $(0,d(a,b))$, and this can only happen if $X$ is uncountable.
Even if useless, this is a pretty result :)
A: Well, I don't have any interesting examples of connected metric spaces which are not immediately obviously uncountable, but here's a stronger theorem:
Every countable regular $T_1$ with at least two points is disconnected.
There are a couple of ways to prove this that I know of; a separation of two arbitrary points can be built successively by taking bigger and bigger disjoint open sets around them with disjoint closures. Alternatively, it can be shown that such a space is necessarily normal, and then Urysohn's lemma can be used to show that any two points are separated.
It is tricky to come up with an example of a countably infinite space which is regular and $T_1$, but not metric (ie, a space which confirms that this theorem really is stronger than the one you gave). I'll withhold my examples for now, in case you decide you want to tackle that problem yourself. As t.b. mentioned in the comments, there are countable Hausdorff connected spaces; that is a more difficult problem to figure out, if you haven't already followed the link.
A: I know of one explicit application of your theorem.  If $X$ is a connected space, suppose there is a cardinal $n \in \lbrace 1, 2, \dots, \aleph_0 \rbrace$ such that for every subset $A \subset X$ with $|A| = n$ we have that $X \setminus A$ is separated.  Then the minimal such cardinal is called the disconnection number of $X$, denoted by $D(X)$.  A compact, connected, metric space for which there exists a disconnection number is called cuttable.
Here is a theorem that uses yours in its proof: If $X$ is cuttable and $D(X) = n$, then for any cardinal $m$ with $n \leq m \leq \aleph_0$ and any set $B \subset X$ with $|B| = m$, we have that $X \setminus B$ is separated.  The proof relies on a theorem called the Boundary Bumping Theorem, which states that in a compact, connected, metric space, if $A \subset X$ and $C$ is a component of $A$, then $\overline{C} \cap \partial(A) \neq \varnothing$.
Proof: Let $|B| = \beta$ and let $A$ be a subset of $B$ with $|A| = \alpha$.  Then $X \setminus A$ is not connected, and by the Boundary Bumping Theorem the closure of each component $C_r$ of $X \setminus A$ intersects $A$.  Picking two components $C_1, C_2$ then since they are non-degenerate, connected metric spaces they're uncountable, and thus neither is contained in $B$ since $|B| = \beta \leq \aleph_0$.  Hence $C_1 \setminus B$ and $C_2 \setminus B$ are non-empty and mutually separated in $X \setminus B$, and thus $B$ also separates $X$.
A concept which your theorem might naturally be concerned with is $\sigma$-connectedness.  A space is $\sigma$-connected if it can't be written as a countable union (other than one containing only a single entry) of pairwise-disjoint closed, connected subsets.  Stone studied these spaces extensively.
A: Call those two points $x_1$ and $x_2$.  So $d(x_1,x_2)>0$.
What is the set $\{d(x_1,x) : x\in A\}$?  Does it contain all numbers between $0$ and $d(x_1,x_2)$?  If so, then you have at least as many points $x\in A$ as numbers between $0$ and $d(x_1,x_2)$.  If not, then some number $c$ between $0$ and $d(x_1,x_2)$ is not the distance between any point $x\in A$ and $x_1$.  So consider the two sets
$$
\{x\in A : d(x_1,x)<c\}\text{ and }\{x\in A : d(x_1,x)>c\}.
$$
Those are disjoint open subsets of $A$ whose union is all of $A$, so $A$ would not be connected.
A: The biggest application that I know for this result is the immediate corollary that a countable metric space is totally disconnected.
A: There may be situations I am unaware of, but I don't think the standard setting is to have a metric space $(X,d)$, where you know $X$ is connected under $d$, and that there are at least two distinct points in $X$, but don't already know the space is uncountable (and care!). I think it would be far more likely to know that $X$ is at most countable, and then we would know the space must be disconnected, regardless of the metric we choose to use.
For example, for those that haven't seen Ostrowski's theorem, and have no idea what metrics can be placed on $\mathbb{Q}$, your result immediately shows it is impossible to construct a metric under which $\mathbb{Q}$ is a connected metric space. (That's not to say it's a bad idea to get your hands dirty, try to build a metric $d'$ so that $(\mathbb{Q},d')$ is a connected metric space, and see what goes wrong!)
One could then see this as an argument to construct $\mathbb{R}$ from $\mathbb{Q}$, since no matter what metric we use, there are holes.
I suppose one could also say, if there is a topology $\tau$ on $\mathbb{Q}$ so that $(\mathbb{Q},\tau)$ is a connected topological space, then we also know that this space is not metrizable. I don't know if this is a particularly useful point of view though..
Of course these are only examples using $\mathbb{Q}$ to illustrate the point, and the same holds for far more odd 'looking' at most countable spaces, where it may be far less intuitive that there are no metrics to make the space a connected metric space.
A: Hint: Find a nonconstant continuous real valued function on M and  Use Very beautiful property of connectedness in M 
Every continuous function preserve connectedness.
Thanks
